What are some vector calculus questions about gradient, divergence, and curl?

[misogynisticfeminist]

1. I was reading on the geometric interpretation of the grad operator. I've did until the point where this particular relation was given.

d\varphi=0=C_1-C_1=\Delta C=(\nabla \varphi)\bullet d \vec r

This is when we permit \vec r to take us from the surface \varphi (x,y,z)=C_1 to another adjacent surface \varphi (x,y,z)=C_2 where c are constants.

Why is it that in the first equation we have C_1-C_1 ?? Also, why is it that the consequence of the first relation shows that,

for a given \vert {d\vec r}\vert, the change of \varphi, d \varphi is maximum when \vert {d\vec r}\vert is parellel to \nabla\varphi when \nabla\varphi is normal to the surface.

2. While evaluating the divergence of vector, \nabla \bullet \vec r f(r) = \frac {\partial}{\partial x} (xf(r))+ \frac {\partial}{\partial y}(yf(r))+ \frac{\partial}{\partial z} (zf(r)),

why is it equivalent to,

3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr} ?

I've tried manipulating the partial differentials using chain rules and all but don't seem to get it. Can someone show me the steps how? Also,

3. I was trying to simplify\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y). Also, how does it reduce to the final answer,

f \nabla \times \vec V \vert _x +\nabla f \times \vec V \vert _x??

I have also tried manipulating the partial derivatives to no avail. Can someone help? thanks alot...

: )

 

[Galileo]

2. While evaluating the divergence of vector, \nabla \bullet \vec r f(r) = \frac {\partial}{\partial x} (xf(r))+ \frac {\partial}{\partial y}(yf(r))+ \frac{\partial}{\partial z} (zf(r)),

why is it equivalent to,

3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr} ?

Here you should use the product and chain rule. Also: r=\sqrt{x^2+y^2+z^2}

So for example:

\frac{\partial}{\partial x}(xf(r))=f(r)+x\frac{\partial}{\partial x}f(r)
This explains where the 3f(r) comes from.
Now:

\frac{\partial}{\partial x}f(r(x,y,z))=\frac{df}{dr}\frac{\partial r}{\partial x}

 

[HallsofIvy]

"This is when we permit to take us from the surface \varphi (x,y,z)=C_1
to another adjacent surface \varphi (x,y,z)=C_1 where c are constants."

No, I don't think that's what that particular formula is intended to show. What that shows is that the derivative in a direction tangent to a level surface is 0 so (\nabla \varphi)\bullet d \vec r= 0: that is, the gradient is perpendicular to all level surfaces.
What is true is that, since \vec u \bullet \vec v= |\vec u||\vec v|cos(\theta)[/tex], (\nabla \varphi)\bullet d \vec r is largest when cos(&amp;theta;) is 1: i.e. &amp;theta;= 0 and the two vectors are parallel.<br /> <br /> For 2, I surprised you didn&#039;t see it while looking at the chain rule.<br /> r= (x^2+ y^2+ z^2)^{\frac{1}{2}} so \frac{\partial r}{\partial x}= \frac{1}{2}(x^2+ y^2+ z^2)^\frac{-1}{2}(2x)= \frac{x}{r}.<br /> Of course, the same is true of the derivative with respect to y and z.<br /> Now, \frac{\partial}{\partial x}(xf(r))= f(r)+ x\frac{\partial f}{/partial x}= f(r)+ \frac{x^2 f&amp;#039;(r)}{r}.<br /> <br /> That first term is what gives you &quot;3f(r)&quot;.

 

[misogynisticfeminist]

Hey, thanks alot, that helped. Actually what i had problems with was the use of the product rule while evaluating \partial/ \ {\partial} x (xf(r)) and i also failed to keep in mind that r=\sqrt {x^2+y^2+z^2 but that was sorted out anyway.

thanks alot. I've still got one question regarding the curl operator though...hope that someone could help me with that.

 

[dextercioby]

In general

(\nabla\times\vec{f})_{x}=...?

Daniel.

 

[misogynisticfeminist]

In general

(\nabla\times\vec{f})_{x}=...?

Daniel.

For, \nabla \times \vec f, it is,

det \left(<br /> \begin{array}{ccc}<br /> <br /> \hat x &amp; \hat y &amp; \hat z\\<br /> {\partial}/{\partial x} &amp; {\partial}/{\partial y} &amp; {\partial}/{\partial z} \\<br /> F_x &amp; F_y &amp; F_z \\<br /> <br /> \end{array}<br /> \right)<br /> <br />

(the matrix is quite horrible though, i can't seem to type it the right way, but i guess you know what i mean).

hmmm, I've tried doing the curl again, i got,

\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)

Using the product rule

( f \frac {\partial{V_z}}{\partial {y}} + {V_z}\frac {\partial {f}}{\partial y})-(f \frac {\partial{V_y}}{\partial {z}} + {V_y}\frac {\partial {f}}{\partial z})

Simplifying,

\frac {\partial}{\partial {y}} (f V_z +V_z f)-\frac {\partial}{\partial {z}} (f V_y+ V_y f)

2f( \frac {\partial}{\partial {y}} V_z - \frac {\partial}{\partial {z}} V_y)

2f \nabla \times \vec {v} \vert _x

Where do i go from here? Or are my steps wrong?

 

[dextercioby]

kay,the code is \begin{array}{ccc} (3 columns)...

Okay.

Now solve your problem.

Daniel.

 

[misogynisticfeminist]

hmmm, I've tried it in the edited post above.

 

[dextercioby]

Well,use the Leibniz rule
[\nabla\times (f\vec{V})]_{x} =[(\nabla f\times \vec{V})+f(\nabla\times\vec{V})]_{x} =...

Daniel.

 

[misogynisticfeminist]

lol, leibniz's rule did everything. But I've arrived at the solution already, thanks alot.